Discover answers to your most pressing questions at Westonci.ca, the ultimate Q&A platform that connects you with expert solutions. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.
Sagot :
The amount of nitrogen gas present at equilibrium is 0.0039 moles.
Volume of the system = 3500 mL or 3.5 L
Concentration of ammonia at equilibrium = 0.25 mol/3.5 L = 0.07 M
Concentration of hydrogen gas at equilibrium = 0.080 mol//3.5 L = 0.02 M
Equilibrium constant for this reaction = 5.81 x 10^5
The reaction for the formation of ammonia is;
N2 + 3H2 ⇄ 2NH3
So;
K = [NH3]^2/[N2] [H2]^3
K [N2] [H2]^3 = [NH3]^2
[N2] = [NH3]^2/K [H2]^3
[N2] = [ 0.07]^2/ 5.81 x 10^5 × [0.02]^3
[N2] = 0.0011 M
Amount of N2 = 0.0011 M × 3.5 L = 0.0039 moles
Learn more: https://brainly.com/question/17960050
Thanks for stopping by. We are committed to providing the best answers for all your questions. See you again soon. Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Thank you for visiting Westonci.ca, your go-to source for reliable answers. Come back soon for more expert insights.